Calculating Emptying Pipe Time

This problem involves calculating the time taken by an emptying pipe based on the filling times of two inlet pipes and the combined filling time when all three pipes are open.

Pipe Rates Calculation

• Let the time taken by the first pipe to fill the cistern be $T_1 = 36$ minutes. Its filling rate is $R_1 = \frac{1}{T_1} = \frac{1}{36}$ cistern per minute.
• Let the time taken by the second pipe to fill the cistern be $T_2 = 45$ minutes. Its filling rate is $R_2 = \frac{1}{T_2} = \frac{1}{45}$ cistern per minute.
• Let the time taken by the bottom pipe to empty the cistern be $T_3$ minutes. Its emptying rate is $R_3 = \frac{1}{T_3}$ cistern per minute. Since it empties, we consider this rate as negative in the combined calculation.
• When all three pipes are open, the cistern fills in $T_{net} = 70$ minutes. The net filling rate is $R_{net} = \frac{1}{70}$ cistern per minute.

Combined Rate Equation

The net rate is the sum of the filling rates minus the emptying rate:

$R_{net} = R_1 + R_2 - R_3$

Substituting the known values:

$\frac{1}{70} = \frac{1}{36} + \frac{1}{45} - R_3$

Solving for the Emptying Rate ($R_3$)

First, find the combined rate of the two filling pipes:

$R_1 + R_2 = \frac{1}{36} + \frac{1}{45}$

The least common multiple (LCM) of 36 and 45 is 180.

$R_1 + R_2 = \frac{5}{180} + \frac{4}{180} = \frac{9}{180} = \frac{1}{20}$ cistern per minute.

Now substitute this back into the net rate equation:

$\frac{1}{70} = \frac{1}{20} - R_3$

Rearrange to solve for $R_3$:

$R_3 = \frac{1}{20} - \frac{1}{70}$

The LCM of 20 and 70 is 140.

$R_3 = \frac{7}{140} - \frac{2}{140} = \frac{5}{140} = \frac{1}{28}$ cistern per minute.

Final Answer: Time to Empty

The time taken by the bottom pipe to empty the completely filled cistern is the reciprocal of its rate $R_3$:

$T_3 = \frac{1}{R_3} = \frac{1}{1/28} = 28$ minutes.